#1
Apr1907, 05:00 AM

P: n/a

In the analysis of sound waves generated in a pipe open at both ends
or open at one end only, the tube length is taken as L=L_0+0.3 D, where L_0 is the actual tube length and D is the diameter of the tube. Is this only a thumb rule? Or can a physical argument be given for the factor 0.3 D? I havent found any argument in undergrad physics books. Can someone help? regards. 


#2
Apr1907, 05:00 AM

P: n/a

vishmz@gmail.com wrote:
> In the analysis of sound waves generated in a pipe open at both ends > or open at one end only, the tube length is taken as L=L_0+0.3 D, > where L_0 is the actual tube length and D is the diameter of the tube. > Is this only a thumb rule? Or can a physical argument be given for the > factor 0.3 D? I havent found any argument in undergrad physics books. > Can someone help? This is a low frequency approximation, that works quite well if the wavelength is smaller than the pipe diameter. At the open end of a pipe, you have velocity fluctuations but little pressure fluctuations. So the impedance here is masslike. The mass impedance of a small slice (dx) of tube is jw rho dx/S. [w is angular frequency, rho is density, S is cross sectional area, and j = the root of 1.] The mass impedance of the open end can be estimated by slicing up space in concentric spherical slices of radius (r). Each slice has mass impedance jw rho dr /(4 pi r^2). Integrate this over r from R (radius of tube) to infinity, and you get jw rho /(4 pi R), the same as a tube slice of length R/4, or D/8. So the above estimate gives an end correction of 0.125 D. But it is a bit crude, since we start integrating from a ball of radius R. In reality, the pattern is more complex. If you account for the pattern in more detail, you get a different value. I don't know if it is 0.3, I thought it was higher. Note that we used low frequency approximation, so that the whole field around the tube rnd is behaving like a mass. It is a bit like an open end has a little ball of air attached to it, whose mass has to be displaced by the acoustic wave. An important application of the end effect is the impedance of a small hole. This is the same as a little pipe of length about 0.8D. You can use this to design all kinds of resonators and dampers, and understand how sound can leak through holes. Gerard 


#3
Apr2007, 05:00 AM

P: n/a

On Apr 19, 3:51 am, Gerard Westendorp <west...@xs4all.nl> wrote:
> vis...@gmail.com wrote: > > In the analysis of sound waves generated in a pipe open at both ends > > or open at one end only, the tube length is taken as L=L_0+0.3 D, > > where L_0 is the actual tube length and D is the diameter of the tube. > > Is this only a thumb rule? Or can a physical argument be given for the > > factor 0.3 D? I havent found any argument in undergrad physics books. > > Can someone help? > > This is a low frequency approximation, that works quite well if the > wavelength is smaller than the pipe diameter. > > At the open end of a pipe, you have velocity fluctuations but little > pressure fluctuations. So the impedance here is masslike. The mass > impedance of a small slice (dx) of tube is jw rho dx/S. [w is angular > frequency, rho is density, S is cross sectional area, and j = the root > of 1.] > > The mass impedance of the open end can be estimated by slicing up space > in concentric spherical slices of radius (r). Each slice has mass > impedance jw rho dr /(4 pi r^2). Integrate this over r from R (radius of > tube) to infinity, and you get jw rho /(4 pi R), the same as a tube > slice of length R/4, or D/8. > > So the above estimate gives an end correction of 0.125 D. But it is a > bit crude, since we start integrating from a ball of radius R. In > reality, the pattern is more complex. If you account for the pattern in > more detail, you get a different value. I don't know if it is 0.3, I > thought it was higher. > > Note that we used low frequency approximation, so that the whole field > around the tube rnd is behaving like a mass. It is a bit like an open > end has a little ball of air attached to it, whose mass has to be > displaced by the acoustic wave. > > An important application of the end effect is the impedance of a small > hole. This is the same as a little pipe of length about 0.8D. You can > use this to design all kinds of resonators and dampers, and understand > how sound can leak through holes. > > Gerard Dear Gerard, Thanx for the explanation!! 


#4
Apr2107, 05:00 AM

P: n/a

End Correction
vishmz@gmail.com wrote:
> On Apr 19, 3:51 am, Gerard Westendorp <west...@xs4all.nl> wrote: ... >> This is a low frequency approximation, that works quite well if the >> wavelength is smaller than the pipe diameter. ... > Thanx for the explanation!! You mean "larger" rather than "smaller"  low frequency corresponds to large wavelength.  Hans Werner Strube strube(@)physik3(.)gwdg(.)de Drittes Physikalisches Institut, Univ. Goettingen FriedrichHundPlatz 1, 37077 Goettingen, Germany 


Register to reply 
Related Discussions  
bolometric correction  Astrophysics  4  
Correction of vision  Introductory Physics Homework  1  
QED correction terms  General Physics  0  
End correction?  Introductory Physics Homework  0  
End Correction  Classical Physics  3 